Theorem oa4b 939

Description: Derivation of 4-OA law variant.

Hypothesis

Ref	Expression
oa4b.1	((a →1 g) ∩ (a ∪ (c ∩ (((a ∩ c) ∪ ((a →1 g) ∩ (c →1 g))) ∪ (((a ∩ e) ∪ ((a →1 g) ∩ (e →1 g))) ∩ ((c ∩ e) ∪ ((c →1 g) ∩ (e →1 g)))))))) ≤ (((a ∩ g) ∪ (c ∩ g)) ∪ (e ∩ g))

Assertion

Ref	Expression
oa4b	((a →1 g) ∩ (a ∪ (c ∩ (((a ∩ c) ∪ ((a →1 g) ∩ (c →1 g))) ∪ (((a ∩ e) ∪ ((a →1 g) ∩ (e →1 g))) ∩ ((c ∩ e) ∪ ((c →1 g) ∩ (e →1 g)))))))) ≤ g

Proof of Theorem oa4b

Step	Hyp	Ref	Expression
1		oa4b.1	. 2 ((a →1 g) ∩ (a ∪ (c ∩ (((a ∩ c) ∪ ((a →1 g) ∩ (c →1 g))) ∪ (((a ∩ e) ∪ ((a →1 g) ∩ (e →1 g))) ∩ ((c ∩ e) ∪ ((c →1 g) ∩ (e →1 g)))))))) ≤ (((a ∩ g) ∪ (c ∩ g)) ∪ (e ∩ g))
2		lear 153	. . . 4 (a ∩ g) ≤ g
3		lear 153	. . . 4 (c ∩ g) ≤ g
4	2, 3	lel2or 162	. . 3 ((a ∩ g) ∪ (c ∩ g)) ≤ g
5		lear 153	. . 3 (e ∩ g) ≤ g
6	4, 5	lel2or 162	. 2 (((a ∩ g) ∪ (c ∩ g)) ∪ (e ∩ g)) ≤ g
7	1, 6	letr 129	1 ((a →1 g) ∩ (a ∪ (c ∩ (((a ∩ c) ∪ ((a →1 g) ∩ (c →1 g))) ∪ (((a ∩ e) ∪ ((a →1 g) ∩ (e →1 g))) ∩ ((c ∩ e) ∪ ((c →1 g) ∩ (e →1 g)))))))) ≤ g

Syntax hints:   ≤ wle 2   ∪ wo 6   ∩ wa 7   →₁ wi1 13

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37

This theorem depends on definitions:  df-a 39  df-t 40  df-f 41  df-le1 122  df-le2 123

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